題目
Given a linked list, remove the nth
node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:Given n will always be valid.Try to do this in one pass.
分析
題目給了一個鏈表的一維數(shù)組辆憔，和一個整數(shù)n怯伊，要求刪除從末尾開始數(shù)第n個節(jié)點灶泵，然后返回鏈表的頭夷野。做法是，用兩個指針澜建，一個fast指針嫉你，一個solw指針九孩，讓slow指針永遠慢fast指針n個節(jié)點。fast走完后阱扬，再用slow指針來刪除需要刪除的節(jié)點
代碼
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* h = new ListNode(0);
h->next = head;
ListNode* slow = h;
ListNode* fast = h;
while(fast->next!=NULL){
if(n<=0){
slow = slow->next;
}
fast= fast->next;
n--;
}
if(slow->next!=NULL){
slow->next = slow->next->next;
}
return h->next;
}
};
